# Self adjoint operators in Kasparov-Modules

In Blackadars book in 17.4.2 it says that for each element $x \in KK(A,B)$ there is a Kasparov module $(E,\pi ,T)$ such that $T=T^*$. Now, the argument for that is that if $(E,\pi, T)$ is any Kasparov-module, then $(E, \pi, (T+T^*)/2)$ is a compact perturbation which would prove the claim. However, I don't see why $(E, \pi, (T+T^*)/2)$ is Kasparov-Module, to be precise I don't see why $((T+T^*)/2)^2 -1$ is compact. It's probably really easy, but I just don't see it at the moment.

Thank you

• Going from memory here, so I might not have this quite right. But as I recall $f(T)$ is compact whenever $f \in C_0(-1,1)$, and $((T + T^*)/2)^2 - 1 = f(T)$ where $f(z) = ((z + \overline{z})/2)^2 - 1 \in C_0(-1,1)$. – Paul Siegel Feb 22 '17 at 21:42
• It is possible that $T^2=1$ but $((T+T^*)/2)^2-1$ is not compact. For example $T=\left\lgroup \matrix{1 & -1\cr 0 & -1} \right\rgroup$ with infinite dimensional blocks. – Caleb Eckhardt Feb 24 '17 at 22:45
• @CalebEckhardt Here "compact" means "compact Kasparov module", so in this case it's something like "$(((T + T^*)/2)^2 - 1)\pi(a)$ is $B$-compact for all $a$". – Paul Siegel Feb 27 '17 at 19:17
• alternatively there is a proof in George Skandalis "Kasparov's bivariant K-theory and applications". In this work he consideres a Kasparov-module $x'=(E,\pi ,T')$ where $T'=T(TT^*)^{\frac{1}{2}}$. And $T_t'=T(TT^*)^{\frac{t}{2}}$ gives the desired homotopy between $x$ and $x'$. – Sabrina Gemsa Apr 27 '17 at 4:46
• and $((\frac{T+T^*}{2})^2-1)\pi(a)= (\frac{(T^2-1)\pi(a)}{2})+(\frac{((T^*)^2-1)\pi(a)}{2})+\frac{(TT^*+T^*T)\pi(a)}{2}$ The first two summands are compact since not only $(E,\pi, T)$ is a Kasparov-module, but also $(E,\pi, T^*)$ is a Kasparov-module. For the third summand maybe you can use $TT^*=(T^*T)^{-1}$? – Sabrina Gemsa Apr 27 '17 at 5:07

What is meant is that $T$ and $(T+T^*)/2$ differ by a compact operator. More precisely: $$\pi(a) (T - (T+T^*)/2)= \pi(a) (T/2 - T^*/2) \equiv \pi(a) (T/2-T/2)=0$$ for all $a \in A$ modulo $K(E)$, because $\pi(a) T \equiv \pi(a) T^*$.